arxiv:1706.04376v1 [math.qa] 14 jun 2017 · 2 liqianbai,xueqingchen,mingdingandfanxu extended the...

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CLUSTER MULTIPLICATION THEOREM IN THE QUANTUM

CLUSTER ALGEBRA OF TYPE A(2)2

LIQIAN BAI, XUEQING CHEN, MING DING AND FAN XU

Abstract. The objective of the present paper is to prove cluster multiplication theorem

in the quantum cluster algebra of type A(2)2 . As corollaries, we obtain bar-invariant

Z[q±1

2 ]-bases established in [6], and naturally deduce the positivity of the elements inthese bases. One bar-invariant basis as the triangular basis of this quantum clusteralgebra is also explicitly described.

1. Introduction

The concept of cluster algebras was introduced by Fomin and Zelevinsky [16][17] in orderto develop an algebraic framework for understanding total positivity and canonical bases insemisimple algebraic groups. As a noncommutative analogue of cluster algebras, quantumcluster algebras were defined by Berenstein and Zelevinsky in [3]. Under the specializationq = 1, the quantum cluster algebras are degenerated to cluster algebras.

For the classical cluster algebras, Sherman and Zelevinsky [26] firstly gave the clustermultiplication formula in rank 2 cluster algebras of finite and affine types. For the generalcase, the cluster categories were introduced in [2] as the categorification of acyclic clusteralgebras. Cluster algebras have a close link to quiver representations via cluster categories.The link is explicitly characterized by the Caldero-Chapoton map [5] and the Caldero-Keller multiplication theorems [7][8]. The Caldero-Chapoton map associates the objectsin the cluster categories to some Laurent polynomials, in particular, sends indecomposablerigid objects to cluster variables. The remarkable Caldero-Keller multiplication theoremsshow the multiplication rules between images of objects under the Caldero-Chapoton map.For simply laced Dynkin quivers, Caldero and Keller constructed a cluster multiplicationformula (of finite type) between two generalized cluster variables in [7]. On the one hand,this multiplication is similar to the multiplication in a dual Hall algebra and unifies ho-mological and geometric properties of cluster categories and combinatorial properties ofcluster algebras. Since cluster algebras were introduced in order to study canonical bases,it is important to construct integral bases of cluster algebras. In the cluster theory, theCaldero-Chapoton map and the Caldero-Keller cluster multiplication theorem open a newway to construct cluster algebras from 2-Calabi-Yau categories and play a very importantrole to obtain structural results such as bases with good properties, positivity conjecture,denominator conjecture and so on [7][13][15]. The cluster multiplication formula of finite

Key words and phrases. quantum cluster algebra, cluster multiplication theorem, positivity.Liqian Bai was supported by NPU (No. 3102017OQD033), Ming Ding was supported by NSF of

China (No. 11771217) and Specialized Research Fund for the Doctoral Program of Higher Education(No. 20130031120004) and Fan Xu was supported by NSF of China (No. 11471177).

1

http://arxiv.org/abs/1706.04376v2

2 LIQIAN BAI, XUEQING CHEN, MING DING AND FAN XU

type was generalized to affine type in [18] and to any type in [27]. Palu [21] further ex-tended the formula to 2-Calabi-Yau categories with cluster tilting objects. In [12], thefull generalization of the Caldero-Chapoton map was obtained for quivers with potentials.Following this link, some good bases have been constructed for finite and affine clusteralgebras [7][9][15].

It is natural to ask the question: what are the quantum analogue of the link? Recently,Rupel [25] defined a quantum version of the Caldero-Chapoton map for the quantum clus-ter algebras over finite fields associated with valued acyclic quivers and he conjectured thatcluster variables could be expressed as images of indecomposable rigid objects under thequantum Caldero-Chapoton formula. A key ingredient of the conjecture is to confirm themutation rules between quantum cluster variables given by [3]. Most recently, the conjec-ture has been proved by Qin [22] for acyclic equally valued quivers. There the author con-structed a quantum cluster multiplication formula and then confirmed the mutation rulesbetween quantum cluster variables. Note that Qin verified the formula for the usual quan-tum cluster algebras through the existence of counting polynomials instead of working overthe finite field. The quantum Caldero-Chapoton maps are further generalized in [11][10].In [14], Ding and Xu proved a multiplication theorem for acyclic quantum cluster algebraswhich generalized the quantum cluster multiplication formula in [22] and could be viewedas a quantum analogue of the one-dimensional Caldero-Keller multiplication theorem dis-cussed in [8]. Compared to the role which the Caldero-Keller multiplication theorems playfor cluster algebras, the quantum multiplication theorem is worthy of highlighting and alsoreflects the information and the difficulty to prove the more general quantum analog of theCaldero-Keller multiplication theorems. By using this multiplication theorem, it is not toodifficult to construct some good ZP-bases in quantum cluster algebras of finite and affinetypes. By specializing q and coefficients to 1, these bases induce the good bases for clusteralgebras of finite [7] and affine types [15], respectively.

One may expect to explicitly express the multiplication of two basis elements in termsof basis elements in quantum cluster algebras, i.e., to get structure constants clearly andexplicitly. Ding and Xu [13] gave the cluster multiplication formula in the quantum clusteralgebra of the Kronecker quiver. By using the multiplication formula, they constructed bar-invariant bases of the quantum cluster algebra of Kronecker quiver as quantum analogues ofthe canonical basis, semicanonical basis and dual semicanonical basis of the correspondingcluster algebra. As a byproduct, they also proved positivity of the elements in these bases.In this paper, we construct a nontrivial cluster multiplication formula in the quantum

cluster algebra of the non-simply-laced valued quiver A(2)2 , which is parallel to the results

obtained for Kornecker quiver but is not a trivial generalization. This formula yields

some important properties of quantum cluster algebra of type A(2)2 . For example, we

construct three integral bases of this quantum cluster algebra. It is worthy of highlightingthat the basis B we obtained coincide with “quantum greedy basis” (or “quantum atomicbasis”, “quantum theta basis”) defined in [20]. In general, quantum greedy basis doesnot have positive structure constants. But it does in this case following from the mainresult Theorem 3.3 of the paper . Whether quantum greedy bases exist or not for generalcases are still not known. The basis S is closely related to the dual canonical basis of aquantum unipotent cell (subset of the dual canonical basis of the Uq(n)), when the valuedquiver is attached some correct frozen variables. In the last section, we prove that S is the

CLUSTER MULTIPLICATION THEOREM IN THE QUANTUM CLUSTER ALGEBRA OF TYPE A(2)2 3

triangular basis in the sense of Qin [23], or the triangular basis in the sense of Berenstein-Zelevinsky [4]. Note that Qin [24] proved that both definitions of the triangular bases areequivalent for the seeds associated with acyclic quivers and for the seeds associated withbipartite skew-symmetrizable matrices. The basis D is similar to the quantum generic basis(quantum dual semi-canonical basis) in [19]. To construct explicit multiplication formulais still open for general cases.

2. Preliminaries

For the terminology related to quantum cluster algebras, one can refer to [3] for more

details, to quantum cluster algebra of type A(2)2 , refer to [6].

In this paper, we consider the valued quiver associated to the compatible pair (Λ, B)where

Λ =

(

0 1−1 0

)

and B =

(

0 1−4 0

)

.

Note that ΛTB =

(

4 00 1

)

.

Now let q denote the formal variable and F be the skew field of fractions of the quantum

torus T = Z[q±12 ]〈X1,X2 | X1X2 = qX2X1〉. The quantum cluster algebra Aq(1, 4) is the

Z[q±12 ]-subalgebra of F generated by the cluster variables Xk for k ∈ Z, recursively defined

by

Xk−1Xk+1 =

{

q12Xk + 1, if k is odd;

q2X4k + 1, if k is even.

Note that Xk ∈ T by the well-known quantum Laurent phenomenon [3]. For each

(a, b) ∈ Z2, if we set X(a,b) = q−

12abXa

1Xb2, then X

(a,b)X(c,d) = q−12(bc−ad)X(a+c,b+d). The

Z-linear bar-involution on the based quantum torus T is defined as follows:

qr2X(a,b) = q−

r2X(a,b), for any r, a and b ∈ Z.

In [6], the authors constructed three kinds of bar-invariant Z[q±12 ]-bases of the quantum

cluster algebra Aq(1, 4) by using the standard monomials discussed in [3]. We now brieflyrecall some notations and results in [6].

We define that

Xδ :=X(−1,−2) +X(−1,2) +X(1,−2) + (q−

12 + q

12 )X(0,−2)

=qX20X3 − q2(qX1 + q−

12 + q

12 )X2

2 .

LetB = {q−

12abXa

mXbm+1 | m ∈ Z, (a, b) ∈ Z

2≥0} ∪ {Fn(Xδ)},

S = {q−12abXa

mXbm+1 | m ∈ Z, (a, b) ∈ Z

2≥0} ∪ {Sn(Xδ)},

D = {q−12abXa

mXbm+1 | m ∈ Z, (a, b) ∈ Z

2≥0} ∪ {Xn

δ },

where Fn(x) and Sn(x) are well-known Chebyshev polynomials defined by

F0(x) = 1, F1(x) = x, F2(x) = x2 − 2, Fn+1(x) = Fn(x)x− Fn−1(x) for n ≥ 2,

S0(x) = 1, S1(x) = x, S2(x) = x2 − 1, Sn+1(x) = Sn(x)x− Sn−1(x) for n ≥ 2,


and Fn(x) = Sn(x) = 0 for n < 0. The homomorphism σ2 : Aq(1, 4) → Aq(1, 4) defined

by Xm 7→ Xm+2 and q±12 7→ q±

12 is an automorphism of Aq(1, 4) [6, Section 4]. Note that

σ2(Xδ) = Xδ.We can define a partial order ≤ on Z

2 as follows: (r1, r2) ≤ (s1, s2) if r1 ≤ s1 andr2 ≤ s2 for (r1, r2), (s1, s2) ∈ Z

2. Moreover if there exists some i ∈ {1, 2} such thatri < si, we will write (r1, r2) < (s1, s2). In [6], the authors showed that every element in{Xn (n ∈ Z \ {1, 2}), Fn(Xδ)(n ≥ 1), Sn(Xδ)(n ≥ 1)} has a minimal non-zero term X(a,b)

according to the partial order ≤. The vector (−a,−b) associated to this minimal non-zero

term X(a,b) of the corresponding element will be called the denominator vector. Then by

using the standard monomials, they proved that B, S and D are bar-invariant Z[q±12 ]-

bases of the quantum cluster algebra Aq(1, 4). Unfortunately, the structure constants andthe positivity are not presented in this construction. This motivated our study for themultiplication formulas.

3. Cluster multiplication theorem and positive bases

In this section, we mainly prove the cluster multiplication theorem of the quantumcluster algebra Aq(1, 4). Since the element Xδ stated in previous section plays a crucialimportance in the cluster multiplication theorem, we firstly address another expression ofthis element.

Lemma 3.1. In Aq(1, 4), we have that Xδ = q−1X24X1 − q−2(q−1X3 + q−

12 + q

12 )X2

2 .

Proof. Note that X3 = X(−1,4) +X(−1,0) and X4 = X(−1,3) +X(−1,−1) +X(0,−1), then wehave that

q−1X24X1 − q−2(q−1X3 + q−

12 + q

12 )X2

2

=q−1(X(−1,3) +X(−1,−1) +X(0,−1))2X(1,0) − (q−3(X3 + q12 + q

32 )X2

2 )

=q−4X(−1,6) +X(−1,−2) +X(1,−2) + (q−12 + q

12 )q−2X(0,2)

+ (q−2 + q2)q−2X(−1,2) + (q−12 + q

12 )X(0,−2)

− (q−4X(−1,6) + q−4X(−1,2) + (q−52 + q−

32 )X(0,2))

=X(−1,−2) +X(1,−2) +X(−1,2) + (q−12 + q

12 )X(0,−2) = Xδ.

�

The following proposition is a special case discussed in [1], here we give an alternativeproof by using the above lemma.

Proposition 3.2. The quantum cluster algebra Aq(1, 4) is the Z[q±12 ]-algebra generated

by {Xm,Xm+1,Xm+2,Xm+3} for any m ∈ Z.

Proof. By the definition of Xδ, we know that Xδ ∈ Z[q±12 ]〈X0,X1,X2,X3〉. We have that

Xδ ∈ Z[q±12 ]〈X1,X2,X3,X4〉 by Lemma 3.1. Then through the automorphism σ2, we can

deduce that Xδ ∈ Z[q±12 ]〈Xm,Xm+1,Xm+2,Xm+3〉 for any m ∈ Z.

Note that for any n ∈ Z, we have that X2nXδ = q−12X2n−2 + q

12X2n+2 (see [6, Propo-

sition 4.2]). Then we can deduce that X2n ∈ Z[q±12 ]〈Xm,Xm+1,Xm+2,Xm+3〉 for any


n ∈ Z. Since X2n−2X2n = q12X2n−1 + 1, we obtain that all cluster variables belong to

Z[q±12 ]〈Xm,Xm+1,Xm+2,Xm+3〉. Thus Aq(1, 4) = Z[q±

12 ]〈Xm,Xm+1,Xm+2,Xm+3〉. �

For each n ∈ Z, we denote by

〈n〉 =

{

1, if n is odd;

2, if n is even.

Let x ∈ R, we have the floor function ⌊x⌋ := max{m ∈ Z | m ≤ x} and the ceilingfunction ⌈x⌉ := min{m ∈ Z | m ≥ x}.

For any m > n ≥ 1, it is easy to show that (see [6, Proposition 4.2]):

Fn(Xδ)Fm(Xδ) = Fm+n(Xδ) + Fm−n(Xδ) and Fn(Xδ)Fn(Xδ) = F2n(Xδ) + 2.

The following cluster multiplication theorem is the main result of the present paper.

Theorem 3.3. In Aq(1, 4), we have that

(1) (i) if m is even and n is positive, then

XmFn(Xδ) = q−n2Xm−2n + q

n2Xm+2n. (3.1)

(ii) if m is odd and n is positive, then

XmFn(Xδ) (3.2)

=q−nX〈m−n〉m−n + qnX

〈m+n〉m+n +

∑

k≥1

(

k∑

l=1

(q−4l−12 + q−

4l−32 + q

4l−32 + q

4l−12 ))Fn−2k(Xδ).

(2) if m is even and n is positive, then

XmXm+2n = qn2X

〈m+n〉m+n +

∑

k≥1

(

2k−1∑

l=1

q−n+12

+l)

Fn−2k+1(Xδ). (3.3)

(3) if m is even and n is positive odd, then

Xm−nXm

=∑

1<2k<n

(

min(4k,n−2k)∑

l=1

q−12−k+l

)

Xm−4k +

qn2X3

m− 23n, n ≡ 0 (mod 3)

qn−12 X⌊m− 2

3n⌋X⌈m− 2

3n⌉, otherwise,

(3.4)

and

Xm+nXm

=∑

1<2k<n

(

min(4k,n−2k)∑

l=1

q12+k−l

)

Xm+4k +

q−n2X3

m+ 23n, n ≡ 0 (mod 3)

q−n+12 X⌊m+ 2

3n⌋X⌈m+ 2

3n⌉, otherwise.

(3.5)

(4) if m is odd and n is positive, then

XmXm+2n

=q2nX2〈m+n〉m+n +

n−1∑

k=1

(

4min(k,n−k)∑

l=1

q−12+l)Xm+2n−2k +

∑

k≥1

cn,kF2n−2k(Xδ), (3.6)


where

cn,k =

k∑

i=1

ai(q−2(n−i)−1 + q4k−2(n+i)+1) +

k−1∑

i=1

bi(q−2(n−i) + q4k−2(n+i)) + bkq

−2(n−k)

and aj =j(j−1)

2 , bj =j(j−1)

2 + ⌈ j2⌉ for positive integer j.

Proof. (1) In order to prove (3.1), it suffices to show that

X2Fn(Xδ) = q−n2X2−2n + q

n2X2+2n.

We will prove the claim by induction on n. When n = 1, it follows from [6, Proposition4.2]. When n = 2, we have that

X2F2(Xδ) = X2(X2δ − 2) = q−

12X0Xδ + q

12X4Xδ − 2X2

=(q−1X−2 +X2) + (X2 + qX6)− 2X2 = q−1X−2 + qX6.

Assume that X2Fn(Xδ) = q−n2X2−2n + q

n2X2+2n for n ≥ 2. Then

X2Fn+1(Xδ) = X2Fn(Xδ)Xδ −X2Fn−1(Xδ)

=q−n+12 X−2n + q−

n−12 X4−2n + q

n−12 X2n + q

n+12 X4+2n − q−

n−12 X4−2n − q

n−12 X2n

=q−n+12 X−2n + q

n+12 X4+2n.

To prove (3.2), it suffices to show that

X1Fn(Xδ) (3.7)

=q−nX〈1−n〉1−n + qnX

〈1+n〉1+n +

∑

k≥1

(

k∑

l=1

(q−4l−12 + q−

4l−32 + q

4l−32 + q

4l−12 )

)

Fn−2k(Xδ).

When n = 1, we have that X1Xδ = q−1X20 + qX2

2 by [6, Proposition 4.2]. When n = 2, wehave that

X1F2(Xδ) = X1(X2δ − 2) = (q−1X2

0 + qX22 )Xδ − 2X1

=q−32 (q−

12X−1 + 1) + q−

12 (q

12X1 + 1) + q

12 (q−

12X1 + 1) + q

32 (q

12X3 + 1)− 2X1

=q−2X−1 + q2X3 + (q−32 + q−

12 + q

12 + q

32 ).


Assume that (3.7) is true. Note that X1Fn+1(Xδ) = X1Fn(Xδ)Xδ −X1Fn−1(Xδ). Whenn is even,

X1Fn+1(Xδ)

=[q−nX1−n + qnX1+n +∑

k≥1

(k

∑

l=1

(q−4l−12 + q−

4l−32 + q

4l−32 + q

4l−12 )Fn−2k(Xδ))]Xδ

− q1−nX22−n − qn−1X2

n −∑

k≥1

(

k∑

l=1

(q−4l−12 + q−

4l−32 + q

4l−32 + q

4l−12 )Fn−1−2k(Xδ))

=q−n−1X2−n + qn+1X2

n+2 +

n2−1

∑

k=1

k∑

l=1

(q−4l−12 + q−

4l−32 + q

4l−32 + q

4l−12 )Fn+1−2k(Xδ)

+

n2

∑

l=1

(q−4l−1

2 + q−4l−32 + q

4l−32 + q

4l−12 )Xδ

=q−n−1X2−n + qn+1X2

n+2 +

n2

∑

k=1

k∑

l=1

(q−4l−12 + q−

4l−32 + q

4l−32 + q

4l−12 )Fn+1−2k(Xδ).

The proof for the odd n is similar.(2) For n ≥ 0, it suffices to show that

X2X2+2n = qn2X<2+n>

2+n +∑

k≥1

(2k−1∑

l=1

q−n+12

+l)Fn−2k+1(Xδ). (3.8)

When n = 1, it is the exchange relation. When n = 2, we have that

X2X6 =q− 1

2X2X4Xδ − q−1X22 = q−

12 (q

12X3 + 1)Xδ − q−1X2

2

=X3Xδ + q−12Xδ − q−1X2

2 = qX24 + q−

12Xδ .

Assume that (3.8) is true. Now we calculate X2X4+2n. Note that

X2+2nXδ = q−12X2n + q

12X2n+4,

we have X2n+4 = q−12X2+2nXδ − q−1X2n.

When n is even, it follows that

X2X4+2n = q−12X2X2+nXδ − q−1X2X2n

=qn−12 X2

2+nXδ +∑

k≥1

(

2k−1∑

l=1

q−n2−1+l)Fn+1−2k(Xδ)Xδ − q

n−32 Xn+1

−∑

k≥1

(

2k−1∑

l=1

q−n2−1+l)Fn−2k(Xδ).

Note that X22+nXδ = q−1Xn+1 + qXn+3 + (q

12 + q−

12 ).


Therefore, we have that

X2X4+2n

=qn−32 Xn+1 + q

n+12 Xn+3 + q

n−12 (q−

12 + q

12 ) +

∑

k≥1

(2k−1∑

l=1

q−n2−1+l)Fn+1−2k(Xδ)Xδ

− qn−32 Xn+1 −

∑

k≥1

(

2k−1∑

l=1

q−n2−1+l)Fn−2k(Xδ)

=qn+12 Xn+3 + (q

n2−1 + q

n2 ) +

n2−1

∑

k=1

(

2k−1∑

l=1

q−n2−1+l)Fn+2−2k(Xδ) +

n−1∑

l=1

q−n2−1+lX2

δ

−n−1∑

l=1

q−n2−1+l

=qn+12 Xn+3 +

n2+1

∑

k=1

(2k−1∑

l=1

q−n2−1+l)Fn+2−2k(Xδ).

Similarly, we can prove the statement for odd n.(3) To prove (3.4), it suffices to show that for a positive odd integer n, we have that

X1X1+n

=∑

1<2k<n

(

min(4k,n−2k)∑

l=1

q−12−k+l

)

Xn+1−4k +

qn2X3

1+n3, n ≡ 0 (mod 3);

qn−12 X⌊1+n

3⌋X⌈1+n

3⌉, otherwise.

(3.9)

When n = 1, it is trivial. When n = 3, note that X4 = q−12X2Xδ − q−1X0 by (3.1). It

follows that

X1X4 = q−12X1X2Xδ − q−1X1X0 = q

12X2X1Xδ − q−1X1X0

=q12X2(q

−1X20 + qX2

2 )− q−1X1X0 = q−12X0 + q

32X3

2 .

Since X4Xδ = q−12X2 + q

12X6 and X6 = q−

12X4Xδ − q−1X2, when n = 5, we have that

X1X6 = X1(q− 1

2X4Xδ − q−1X2) = q−1X0Xδ + qX32Xδ − q−1X1X2

=q−32X−2 + q−

12X2 + q

12X2

2X0 + q32X2

2X4 − q−1X1X2

=q−32X−2 + (q−

12 + q

12 + q

32 )X2 + q2X2X3.

Assume that (3.9) is true. Note that X1X3+n = q−12X1X1+nXδ − q−1X1Xn−1.

If n ≡ 0 (mod 3), then

X1Xn+1 =∑

1<2k<n

(

min(4k,n−4k)∑

l=1

q−12−k+l)Xn+1−4k + q

n2X3

1+n3


and

q−1X1Xn−1 =∑

1<2k<n−2

(

min(4k,n−2−2k)∑

l=1

q−32−k+l)Xn−1−4k + q

n−52 Xn

3X1+n

3.

We then get

q−12X1Xn+1Xδ

=∑

1<2k<n

(

min(4k,n−2k)∑

l=1

q−1−k+l)(q−12Xn−1−4k + q

12Xn+3−4k)

+ qn−12 X2

1+n3(q−

12Xn

3−1 + q

12X3+n

3)

=∑

1<2k<n

(

min(4k,n−2k)∑

l=1

q−32−k+l)Xn−1−4k +

∑

1<2k<n

(

min(4k,n−2k)∑

l=1

q−12−k+l)Xn+3−4k

+ qn−32 X1+n

3Xn

3+ q

n+12 X1+n

3X2+n

3+ (q

n−22 + q

n2 )X1+n

3.

Note that ⌊n−26 ⌋ = n−3

6 , ⌈n−26 ⌉ = n+3

6 , ⌊n6 ⌋ = n−36 , ⌈n6 ⌉ = n+3

6 , ⌊n+26 ⌋ = n−3

6 and

⌈n+26 ⌉ = n+3

6 since n ≡ 0 (mod 3). Then we have

4k < n− 2− 2k, if 1 ≤ k ≤n− 3

6,

4k > n− 2− 2k, ifn + 3

6≤ k ≤

n− 3

2,

4k < n− 2k, if 1 ≤ k ≤n− 3

6,

4k > n− 2k, ifn + 3

6≤ k ≤

n− 1

2,

4k < n+ 2− 2k, if 1 ≤ k ≤n− 3

6,

4k > n+ 2− 2k, ifn + 3

6≤ k ≤

n + 1

2.

(3.10)

It follows that

∑

1<2k<n

(

min(4k,n−2k)∑

l=1

q−32−k+l)Xn−1−4k −

∑

1<2k<n−2

(


l=1

q−32−k+l)Xn−1−4k

=

n−32

∑

k=n−36

(n−2k∑

l=n−1−2k

q−32−k+l)Xn−1−4k + q−

n2X1−n − (q

n−22 + q

n2 )X1+n

3.


Hence

X1Xn+3 =

n−12

∑

k=n+36

(

n+2−2k∑

l=n+1−2k

q−12−k+l)Xn+3−4k + q−

n2X1−n

+∑

1<2k<n

(

min(4k,n−2k)∑

l=1

q−12−k+l)Xn+3−4k + q

n+12 X1+n

3X2+n

3

=∑

1<2k<n+2

(

min(4k,n+2−2k)∑

l=1

q−12−k+l)Xn+3−4k + q

n+12 X1+n

3X2+n

3.


X1Xn+1 =∑

1<2k<n

(

min(4k,n−2k)∑

l=1

q−12−k+l)Xn+1−4k + q

n−12 Xn+2

3Xn+5

3,

q−1X1Xn−1 =∑

1<2k<n−2

(


l=1

q−32−k+l)Xn−1−4k + q

n−52 Xn−1

3Xn+2

3.

It follows that

q−12X1Xn+1Xδ

=∑

1<2k<n

(

min(4k,n−2k)∑

l=1

q−32−k+l)Xn−1−4k +

∑

1<2k<n

(

min(4k,n−2k)∑

l=1

q−12−k+l)Xn+3−4k

+ qn−32 Xn+2

3Xn−1

3+ q

n−22 Xn−1

3+ q

n+22 X3

n+53

.

Note that ⌊n−26 ⌋ = n−7

6 , ⌈n−26 ⌉ = n−1

6 , ⌊n6 ⌋ = n−16 , ⌈n6 ⌉ = n+5

6 , ⌊n+26 ⌋ = n−1

6 and

⌈n+26 ⌉ = n+5

6 , therefore

4k < n− 2− 2k, if 1 ≤ k ≤n− 7

6,

4k > n− 2− 2k, ifn− 1

6≤ k ≤

n− 3

2,

4k < n− 2k, if 1 ≤ k ≤n− 1

6,

4k > n− 2k, ifn+ 5

6≤ k ≤

n− 1

2,

4k < n+ 2− 2k, if 1 ≤ k ≤n− 1

6,

4k > n+ 2− 2k, ifn+ 5

6≤ k ≤

n+ 1

2.

(3.11)


It follows that

∑

1<2k<n

(

min(4k,n−2k)∑

l=1

q−32−k+l)Xn−1−4k −

∑

1<2k<n−2

(


l=1

q−32−k+l)Xn−1−4k

=

n−32

∑

k=n+56

(

n−2k∑

l=n−1−2k

q−32−k+l)Xn−1−4k + q

n2−2Xn−1

3+ q−

n2X1−n

=

n−32

∑

k=n−16

(n−2k∑

l=n−1−2k

q−32−k+l)Xn−1−4k + q−

n2X1−n − q

n2−1Xn−1

3.

Hence

X1Xn+3 =

n−12

∑

k=n+56

(

n+2−2k∑

l=n+1−2k

q−12−k+l)Xn+3−4k + q−

n2X1−n

+∑

1<2k<n

(

min(4k,n−2k)∑

l=1

q−12−k+l)Xn+3−4k + q

n+22 X3

n+53

=∑

1<2k<n

(

min(4k,n+2−2k)∑

l=1

q−12−k+l)Xn+3−4k + q−

n2X1−n + q

n+22 X3

n+53

=∑

1<2k<n+2

(

min(4k,n+2−2k)∑

l=1

q−12−k+l)Xn+3−4k + q

n+22 X3

n+53

.


X1Xn+1 =∑

1<2k<n

(

min(4k,n−2k)∑

l=1

q−12−k+l)Xn+1−4k + q

n−12 Xn+1

3Xn+4

3,

and

q−1X1Xn−1 =∑

1<2k<n−2

(


l=1

q−32−k+l)Xn−1−4k + q

n−42 X3

n+13

.

Note that

q−12X1X1+nXδ

=∑

1<2k<n

(

min(4k,n−2k)∑

l=1

q−32−k+l)Xn−1−4k +

∑

1<2k<n

(

min(4k,n−2k)∑

l=1

q−12−k+l)Xn+3−4k

+ qn−42 X3

n+13

+ qn+12 Xn+4

3Xn+7

3+ q

n2Xn+7

3.


Since ⌊n−26 ⌋ = n−5

6 , ⌈n−26 ⌉ = n+1

6 , ⌊n6 ⌋ = n−56 , ⌈n6 ⌉ = n+1

6 , ⌊n+26 ⌋ = n+1

6 , ⌈n+26 ⌉ = n+7

6 ,we have that

4k < n− 2− 2k, if 1 ≤ k ≤n− 5

6,

4k > n− 2− 2k, ifn+ 1

6≤ k ≤

n− 3

2,

4k < n− 2k, if 1 ≤ k ≤n− 5

6,

4k > n− 2k, ifn+ 1

6≤ k ≤

n− 1

2,

4k < n+ 2− 2k, if 1 ≤ k ≤n+ 1

6,

4k > n+ 2− 2k, ifn+ 7

6≤ k ≤

n+ 1

2.

(3.12)

Note that

∑

1<2k<n

(

min(4k,n−2k)∑

l=1

q−32−k+l)Xn−1−4k −

∑

1<2k<n−2

(


l=1

q−32−k+l)Xn−1−4k

=

n−32

∑

k=n+16

(n−2k∑

l=n−2k−1

q−32−k+l)Xn−1−4k + q−

n2X1−n

and2n+2

3∑

l=1

q−12−n+1

6+lXn+7

3−

2n−13

∑

l=1

q−12−n+1

6+lXn+7

3= q

n2Xn+7

3.

It follows that

X1Xn+3

=

n−12

∑

k=n+76

(n+2−2k∑

l=n+1−2k

q−12−k+l)Xn+3−4k + q−

n2X1−n

+∑

1<2k<n

(

min(4k,n−2k)∑

l=1

q−12−k+l)Xn+3−4k + q

n2Xn+7

3+ q

n+12 Xn+4

3Xn+7

3

=∑

1<2k<n+2

(

min(4k,n+2−2k)∑

l=1

q−12−k+l)Xn+3−4k + q

n+12 Xn+4

3Xn+7

3.

This completes the proof of (3.4). The proof of (3.5) is similar to the proof of (3.4).(4) To prove (3.6), it suffices to prove that

X1X1+2n = q2nX2〈1+n〉1+n +

n−1∑

k=1

(

4min(k,n−k)∑

l=1

q−12+l)X2n−2k+1 +

∑

k≥1

cn,kF2n−2k(Xδ). (3.13)


When n = 1, we have that X1X3 = q2X42 + 1 which is the exchange relation. When

n = 2, note that X3F2(Xδ) = q−2X1 + q2X5 + (q−32 + q−

12 + q

12 + q

32 ) by (3.2), then we

have that

X1X5 = q−2X1X3F2(Xδ)− q−4X21 − (q−

72 + q−

52 + q−

32 + q−

12 )X1

=q−1X32X−2 + qX3

2X6 + q−2F2(Xδ)− q−4X21 − (q−

72 + q−

52 + q−

32 + q−

12 )X1.

Note that

q−1X32X−2 = q−2X2

2X20 + q−

12X2

2Xδ ,

qX32X6 = q2X2

2X24 + q

12X2

2Xδ ,

X22X

20 = q−

12X2X1X0 +X2X0 = q−2X2

1 + (q−32 + q−

12 )X1 + 1,

X22Xδ = X2(q

− 12X0 + q

12X4) = q−1X1 + qX3 + (q−

12 + q

12 ),

X22X

24 = q

12X2X3X4 +X2X4 = q2X2

3 + (q12 + q

32 )X3 + 1.

It follows that

X1X5 = q4X23 + (q

12 + q

32 + q

52 + q

72 )X3 + q−2F2(Xδ) + (q−2 + q−1 + 2 + q + q2).

Assume that (3.13) is true. We have that

X1X3+2n = q−2X1X1+2nF2(Xδ)− q−4X1X2n−1 − (q−72 + q−

52 + q−

32 + q−

12 )X1.

If n is even, then

q−2X1X1+2nF2(Xδ)

=q2n−2X21+nF2(Xδ) +

n−1∑

k=1

(

4min(k,n−k)∑

l=1

q−52+l)X2n+1−2kF2(Xδ)

+∑

k≥1

q−2cn,kF2n−2k(Xδ)F2(Xδ).

For convenience, we set

A =q2n−2X21+nF2(Xδ),

B =

n−1∑

k=1

(

4min(k,n−k)∑

l=1

q−52+l)X2n+1−2kF2(Xδ),

C =∑

k≥1

q−2cn,kF2n−2k(Xδ)F2(Xδ),

D =q−4X1X2n−1,

E =(q−72 + q−

52 + q−

32 + q−

12 )X1.

Then we have

A = q2n−2X1+n(q−2Xn−1 + q2Xn+3 + (q−

32 + q−

12 + q

12 + q

32 ))

=q2n−6X4n + q2n+2X4

n+2 + q2n−4 + q2n + (q2n−72 + q2n−

52 + q2n−

32 + q2n−

12 )Xn+1,


B =

n−1∑

k=1

4min(k,n−k)∑

l=1

q−52+l(q−2X2n−1−2k + q2X2n+3−2k + q−

32 + q−

12 + q

12 + q

32 )

=

n−1∑

k=1

(

4min(k,n−k)∑

l=1

q−92+l)X2n−1−2k +

n−1∑

k=1

(

4min(k,n−k)∑

l=1

q−12+l)X2n+3−2k

+n−1∑

k=1

4min(k,n−k)∑

l=1

(q−4+l + q−3+l + q−2+l + q−1+l),

C =

n−2∑

k=1

cn+1,k[F2n−2−2k(Xδ) + F2n+2−2k(Xδ)] + cn+1,n−1(F4(Xδ) + 2) + cn+1,nF2(Xδ)

and

D = q2n−6X4n +

n−2∑

k=1

(

min(k,n−1−k)∑

l=1

q−92+l)X2n−1−2k +

∑

k≥1

cn+1,kF2n−2−2k(Xδ).

It follows that

X1X3+2n = A+B + C −D − E

=q2n−4 + q2n + q2n+2X4n+2 + (q2n−

72 + q2n−

52 + q2n−

32 + q2n−

12 )Xn+1

+n−1∑

k=1

(

4min(k,n−k)∑

l=1

q−92+l)X2n−1−2k +

n−1∑

k=1

(

4min(k,n−k)∑

l=1

q−12+l)X2n+3−2k

+

n−1∑

k=1

4min(k,n−k)∑

l=1

(q−4+l + q−3+l + q−2+l + q−1+l) +

n−2∑

k=1

cn+1,kF2n+2−2k(Xδ)

+ cn+1,n−1(F4(Xδ) + 1) + cn+1,nF2(Xδ)−

n−2∑

k=1

(

4min(k,n−1−k)∑

l=1

q−92+l)X2n−1−2k

− (q−72 + q−

52 + q−

32 + q−

12 )X1

=q2n−4 + q2n + q2n+2X4n+2 + (q2n−

72 + q2n−

52 + q2n−

32 + q2n−

12 )Xn+1

+

n−2∑

k=1

(

4min(k,n−k)∑

l=1

q−92+l)X2n−1−2k +

n−1∑

k=1

(

4min(k,n−k)∑

l=1

q−12+l)X2n+3−2k

+

n−1∑

k=1

4min(k,n−k)∑

l=1

(q−4+l + q−3+l + q−2+l + q−1+l) +

n∑

k=1


+ cn+1,n−1 −n−2∑

k=1

(


l=1

q−92+l)X2n−1−2k.


Since

min(k, n− 1− k) =

k, if k ≤n

2− 1;

n− 1− k, if k ≥n

2,

min(k, n − k) =

k, if k ≤n

2− 1;

n− k, if k ≥n

2,

and

min(k, n+ 1− k) =

k, if k ≤n

2;

n+ 1− k, if k ≥n

2+ 1,

we have that

n−2∑

k=1

(

4min(k,n−k)∑

l=1

q−92+l)X2n−1−2k −

n−2∑

k=1

(


l=1

q−92+l)X2n−1−2k

=n−2∑

k=n2

(

4(n−k)∑

l=1

q−92+l)X2n−1−2k −

n−2∑

k=n2

(

4(n−1−k)∑

l=1

q−92+l)X2n−1−2k

=

n−2∑

k=n2

(

4(n−k)∑

l=4(n−k)−3

q−92+l)X2n−1−2k.

Note that

(q2n−72 + q2n−

52 + q2n−

32 + q2n−

12 )Xn+1 +

n−2∑

k=n2

(

4(n−k)∑

l=4(n−k)−3

q−92+l)X2n−1−2k

+n−1∑

k=1

(

4min(k,n−k)∑

l=1

q−12+l)X2n+3−2k

=

n∑

k=n2+1

(

4(n+1−k)∑

l=4(n−k)+1

q−12+l)X2n+3−2k +

n−1∑

k=1

(

4min(k,n−k)∑

l=1

q−12+l)X2n+3−2k

=n∑

k=1

(

4min(k,n+1−k)∑

l=1

q−12+l)X2n+3−2k.


Therefore

X1X3+2n

=q2n+2X4n+2 +

n∑

k=1

(

4min(k,n+1−k)∑

l=1

q−12+l)X2n+3−2k +

n∑

k=1


+ q2n−4 + q2n +n−1∑

k=1

4min(k,n−k)∑

l=1

(q−4+l + q−3+l + q−2+l + q−1+l) + cn+1,n−1.

We need to show that

q2n−4 + q2n +

n−1∑

k=1

4min(k,n−k)∑

l=1

(q−4+l + q−3+l + q−2+l + q−1+l) + cn+1,n−1 = cn+1,n+1.

(3.14)

Note that cn+1,n−1 =n−1∑

i=1ai(q

−2(n−i)−3+q2(n−i)−5)+n−2∑

i=1bi(q

−2(n−i)−2+q2(n−i)−6)+bn−1q−4.

Let Pn :=n−1∑

k=1

4min(k,n−k)∑

l=1

(q−4+l + q−3+l + q−2+l + q−1+l).

We consider the coefficients of qx in the left hand side of (3.14) for 0 ≤ x ≤ 2n. Whenn = 2, we have that c3,1 = q−4 and P2 = q−3 + 2q−2 + 3q−1 + 4 + 3q + 2q2 + q3. Thusc3,3 = 1 + q4 + c3,1 + P2.

When n = 4, c5,3 = q−8 + q−7 + 2q−6 + 3q−5 + 5q−4 + 3q−3 + 2q−2 + q−1 + 1 and P4 =3q−3+6q−2+9q−1+12+10q+8q2+6q3+4q4+3q5+2q6+q7. Hence c5,5 = q4+q8+c5,3+P4.

Let n ≥ 6, we have that

(i) when x ≥ 2n − 7, the coefficients of qx in cn+1,n−1 are 0. The coefficients of qx inPn are 0 for x ≥ 2n. For 0 ≤ x ≤ 2n − 4, the coefficients of qx in the polynomialPn are 4n− 4− 2x;

(ii) it is easy to see that the coefficients of q2n−4, q2n−3, q2n−2, q2n−1 and q2n in Pn are4, 3, 2, 1 and 0, respectively. Thus the coefficients of q2n−4, q2n−3, q2n−2, q2n−1 andq2n in the left hand side of (3.14) are 5, 3, 2, 1 and 1, respectively;

(iii) when 0 ≤ x ≤ 2n − 8 and x ≡ 0 (mod 4), the coefficients of qx in cn+1,n−1 arebn−3−x

2, then the coefficients of qx in the left hand side of (3.14) are bn−3−x

2+

(4n − 4 − 2x) = bn+1−x2since bn−3−x

2= 1

2(n2 − (6 + x)n + (14x

2 + 3x + 10)) and

bn+1−x2= 1

2(n2 + (2− x)n+ (14x

2 − x+ 2));

(iv) when 0 ≤ x ≤ 2n − 8 and x ≡ 2 (mod 4), the coefficients of qx in cn+1,n−1 arebn−3−x

2, then the coefficients of qx in the left hand side of (3.14) are bn−3−x

2+

(4n − 4 − 2x) = bn+1−x2since bn−3−x

2= 1

2 (n2 − (6 + x)n) + (14x

2 + 3x + 9) and

bn+1−x2= 1

2(n2 + (2− x)n+ (14x

2 − x+ 1));

(v) when 1 ≤ x ≤ 2n− 7 and x are odd, the coefficients of qx in cn+1,n−1 are an−x+52,

so the coefficients of qx in the left hand side of (3.14) are an−x+52

+(4n− 4− 2x) =12(n

2−(x+6)n+(14x2+3x+ 35

4 ))+(4n−4−2x) = 12(n

2+(2−x)n+(14x2−x+ 3

4)) =an+ 3−x

2;


(vi) the coefficient of q2n−6 in Pn and cn+1,n−1 are 8 and 0 respectively. It follows thatthe coefficient of q2n−6 in the left hand side of (3.14) is b4 = 8. The coefficient ofq2n−5 in Pn and cn+1,n−1 are 6 and 0 respectively. Thus the coefficient of q2n−5 inthe left hand side of (3.14) is a4 = 6.

It follows that the coefficients of qx in the left hand side of (3.14) are{

bn+1−x2, if x ∈ {0, 2, . . . , 2n},

an−x−32, if x ∈ {1, 3, . . . , 2n − 1}.

We consider the coefficients of q−x for 1 ≤ x ≤ 2n. We have that

(i) when 4 ≤ x ≤ 2n, the coefficients of q−x in the polynomial Pn are 0;(ii) when x = 1, 2, 3, the coefficients of q−1, q−2, q−3 in the polynomial Pn are 3(n− 1),

2(n−1) and n−1 respectively. Note that the coefficients of q−1, q−2, q−3 in cn+1,n−1

are an−2, bn−2 and an−1 respectively. Hence the coefficients of q−1, q−2, q−3 in theleft hand side of (3.14) are an+1, bn and an respectively;

(iii) when x = 4, 6, 8, . . . , 2n, the coefficients of q−x in cn+1,n−1 are bn+1−x2. Thus the

coefficients of q−x in the left hand side of (3.14) are bn+1−x2for x = 4, 6, 8, . . . , 2n;

(iv) when x = 5, 7, . . . , 2n − 1, the coefficients of q−x in cn+1,n−1 are an−x−32. We

obtain that the coefficients of q−x in the left hand side of (3.14) are an−x−32

for

x = 5, 7, . . . , 2n − 1.

Hence the coefficients of q−x in the left hand side of (3.14) are{

bn+1−x2, if x ∈ {2, . . . , 2n},

an−x−32, if x ∈ {1, 3, . . . , 2n − 1}.

The identity (3.14) is true. This proceeds the induction and finishes the proof.If n is odd, by a similar detailed calculation, we can proceed the proof. Thus we can

draw the conclusion that (3.6) is true.The entire proof of the theorem is now completed. �

Remark 3.4. According to the definition of bar-invariant and Theorem 3.3, we can easilyobtain the similar cluster multiplication formulas for Fn(Xδ)Xm and XnXm in other cases.

As an immediate consequence of the above theorem, we can naturally obtain the follow-ing result proved in [6] by a simple way.

Corollary 3.5. The sets B, S and D are bar-invariant Z[q±12 ]-bases of Aq(1, 4).

Proof. Since there exist unipotent transformations between Fn(Xδ), Sn(Xδ) and Xnδ for

n ≥ 1, we only need to prove that the set B is a bar-invariant Z[q±12 ]-basis of Aq(1, 4).

It suffices to show that B are linearly independent. Note that the denominator vectors ofFn(Xδ) in Aq(1, 4) are (n, 2n) for n ∈ N which are bijective to the set of positive imaginaryroots of the corresponding Lie algebra denoted by Φim

+ . By [26, Proposition 3.1], thereexists a bijection between the set of all denominator vectors of cluster monomials andQ − Φim

+ with Q := Z2 being a lattice of rank 2 with a fixed basis of two simple roots

{α1, α2}. Thus the denominator vectors of cluster monomials and Fn(Xδ) are different


from each other. Assume that S is a finite set and∑

α∈S

aαXα = 0

for Xα ∈ B and aα ∈ Z[q±12 ] \ {0}. Let V (S) denote the set of denominator vectors of Xα

for α ∈ S and let ≤ be the partial order on V (S) inherited from Z2. There exists β ∈ S such

that the denominator vector of Xβ is a maximal denominator vector in (V (S),≤). We canthen deduce that aβ = 0 which is a contradiction. Therefore B is linearly independent. �

Recall that an element Y ∈ Aq(1, 4) is positive if the coefficients in its Laurent expansion

in the cluster variables from {Xm,Xm+1} belong to N[q±12 ]. The following result can be

deduced from Theorem 3.3.

Corollary 3.6. The elements in the Z[q±12 ]-bases B, S and D of Aq(1, 4) are positive.

Proof. By the definitions of Chebyshev polynomials Fn(x) and Sn(x), we only need toprove the positivity of the elements in B. Using the fact that σ2 is an automorphism, itsuffices to prove the positivity in the clusters {X1,X2} and {X2,X3}. Note that

X0 =X(1,−1) +X(0,−1),

X−1 =X(3,−4) + (q−

32 + q−

12 + q

12 + q

32 )X(2,−4) + (q−2 + q−1 + 1 + q + q2)X(1,−4)

+ (q−32 + q−

12 + q

12 + q

32 )X(0,−4) +X(−1,0) +X(−1−4),

X−2 =X(−1,1) +X(2,−3) + (q−1 + 1 + q)X(1,−3) + (q−1 + 1 + q)X(0,−3) +X(−1,−3),

Xδ =X(−1,−2) +X(−1,2) +X(1,−2) + (q−

12 + q

12 )X(0,−2),

F2(Xδ) =X(−2,−4) + (q−2 + q2)X(−2,0) + (q−2 + q−1 + 2 + q + q2)X(0,−4)

+ (q−32 + q−

12 + q

12 + q

32 )(X(−1,4) +X(1,−4) +X(−1,0)) +X(−2,4) +X(2,−4).

Hence X−2,X−1,X0,X1,X2,Xδ and F2(Xδ) are positive elements in {X1,X2}. SupposethatX−n,X−n+1, . . . ,X−2,X−1,X0,X1,X2, . . . ,Xn−1,Xn,Xδ, F2(Xδ), . . . , and Fn(Xδ) arepositive. By Theorem 3.3, when n is odd, we have

X1X1+n =∑

1<2k<n

(

min(4k,n−2k)∑

l=1

q−12−k+l)Xn+1−4k +

qn2X3

1+n3, n ≡ 0 (mod 3);

qn−12 X⌊1+n

3⌋X⌈1+n

3⌉, otherwise,

X1X−n−1

=∑

1<2k<n+2

(

min(4k,n+2−2k)∑

l=1

q12+k−l)X−n−1+4k +

q−n+22 X3

1−n3

, n ≡ 1 (mod 3);

q−n+32 X⌊ 1−n

3⌋X⌈ 1−n

3⌉, otherwise,

X1X2+n = qn+1X2〈n+3

2〉

n+32

+

n−12

∑

k=1

(

4min(k,n+12

−k)∑

l=1

q−12+l)Xn+2−2k +

∑

k≥1

cn+12,kFn+1−2k(Xδ)


and

X−n−2X1 = qn+3X2〈−n+1

2〉

−n+12

+

n+12

∑

k=1

(

4min(k,n+32

−k)∑

l=1

q−12+l)X1−2k +

∑

k≥1

cn+32,kFn+3−2k(Xδ);

when n is even, we have

X1X1+n =qnX2〈n

2+1〉

n2+1 +

n−22

∑

k=1

(

4min(k,n2−k)

∑

l=1

q−12+l)Xn+1−2k +

∑

k≥1

cn2,kFn−2k(Xδ),

X−n−1X1 =qn+2X

2〈−n2〉

−n2

+

n2

∑

k=1

(

4min(k,n2+1−k)

∑

l=1

q−12+l)X1−2k +

∑

k≥1

cn2+1,kFn+2−2k(Xδ),

X2X2+n =qn4X

〈2+n2〉

2+n2

+∑

k≥1

(2k−1∑

l=1

q−n+24

+l)Fn2+1−2k(Xδ),

X−n−2X2 =qn+44 X

〈−n2〉

−n2

+∑

k≥1

(

2k−1∑

l=1

q−n+64

+l)Fn2+3−2k(Xδ).

We deduce that X−n−2,X−n−1,Xn+1 and Xn+2 are positive in {X1,X2}. According toTheorem 3.3, we have that

X1Fn+1(Xδ)

=q−(n+1)X〈−n〉−n + qn+1X

〈n+2〉n+2 +

∑

k≥1

(k

∑

l=1

(q−4l−1

2 + q−4l−32 + q

4l−32 + q

4l−12 ))Fn+1−2k(Xδ).

It follows that Fn+1(Xδ) is positive in {X1,X2}. By induction, each element in B ispositive in {X1,X2}. Similarly, each element in B is positive in {X2,X3}. The proof iscompleted. �

A Z[q±12 ]-basis C of Aq(1, 4) is called the canonical basis of Aq(1, 4) if every positive

element of Aq(1, 4) is the N[q±12 ]-linear combination of elements of C.

Theorem 3.7. The basis B is the canonical basis of Aq(1, 4).

Proof. Let Y be a positive element of Aq(1, 4). Since B is a Z[q±12 ]-basis of Aq(1, 4), we

have Y =∑

Z∈BaZZ for aZ ∈ Z[q±

12 ] \ {0}. It suffices to prove that each aZ equals some

coefficient in the Laurent expansion of Y with respect to Xm and Xm+1 for m ∈ Z.

First of all, we consider the coefficient of the cluster monomial q−12abXa

mXbm+1, where

a, b ∈ N. Since σ2 is an automorphism, it suffices to show that the coefficient aX(c,d) of

X(c,d) in Y =∑

Z∈BaZZ equals the coefficient of X(c,d) in the Laurent expansion of Y with

respect to {X1,X2} (showing that the coefficient of the cluster monomial q−12cdXc

2Xd3 in

Y =∑

Z∈B

aZZ equals the coefficient of q−12cdXc

2Xd3 in the Laurent expansion of Y with


respect to {X2,X3} uses the same method). For m 6= 1, let

q−12abXa

mXbm+1 =

∑

(c,d)∈Z2

Mcd(q)X(c,d)

be the Laurent expansion of the cluster monomial q−12abXa

mXbm+1 with respect to X1 and

X2, where Mcd(q) ∈ Z[q±12 ]. For n ≥ 1, let

Fn(Xδ) =∑

(e,f)∈Z2

Nef (q)X(e,f)

be the Laurent expansion of Fn(Xδ) with respect to X1 and X2, where Nef (q) ∈ Z[q±12 ].

By Corollary 3.6,Mcd(q), Nef (q) ∈ N[q±12 ]. We only need to show that if c, d, e, f ∈ N then

Mcd(q) = Nef (q) = 0. If there exists some Mcd(q) 6= 0 for some c, d ∈ N, then Mcd(1) > 0which contradicts the statement in [26, Proposition 3.6]. Thus Mcd(q) = 0 for c, d ≥ 0. Ifthere exists some Nef (q) 6= 0 for e, f ∈ N, then Nef (1) > 0 which contradicts the statementin [26, Proposition 5.2(2)]. Hence Nef (q) = 0 for e, f ≥ 0.

It remains to consider the coefficient of Fn(Xδ) in Y =∑

Z∈BaZZ for n ≥ 1. Without loss

of generality, we can assume that the cluster variables Xm occurring in Y =∑

Z∈BaZZ satisfy

that m ≥ 3 since σ2 is an automorphism. It suffices to show that, the coefficient of thecluster monomial X(n,−2n) (n ≥ 1) in the Laurent expansion of Fn(Xδ) with respect to the

cluster {X1,X2} is 1, but the coefficients ofX(n,−2n) in the Laurent expansions of Fk(Xδ) or

q−12n1n2Xn1

m Xn2m+1 for k 6= n and m ≥ 3 are 0. Let R1(q) denote the coefficient of X(n,−2n)

in the Laurent expansion of Fk(Xδ) with respect to {X1,X2}. Note that R1(q) ∈ N[q±12 ]

since Fn(Xδ) is positive. Using the fact that Fn(Xδ) is bar-invariant and R1(1) = 1 by [26,

Proposition 5.1], it follows that R1(q) = 1. Let R2(q) ∈ N[q±12 ] denote the coefficient of

X(n,−2n) in the Laurent expansion of Fk(Xδ) with respect to {X1,X2}. If R2(q) 6= 0, then

R2(1) 6= 0, which contradicts [26, Proposition 5.2(1)]. Thus R2(q) = 0. Let R3(q) ∈ N[q±12 ]

denote the coefficient of X(n,−2n) in the Laurent expansion of q−12Xn1

m Xn2m+1 with respect

to {X1,X2}. If R3(q) 6= 0, then R3(1) 6= 0, which contracts the statement in [26, Corollary3.7]. Hence R3(q) = 0. This finishes the proof. �

4. triangular bases

In [4], the choice of the exchange matrix and the skew-symmetric matrix are B =(

0 −bc 0

)

for b, c > 0 and Λ =

(

0 −11 0

)

, respectively. Though our exchange matrix B

and the skew-symmetric matrix Λ differ from those of [4] by a sign, let v = q−12 to reconcile

the formal variables q and v, then Theorem 3.3 is still true.The following theorem is a version of Lusztig’s Lemma.

Theorem 4.1. [4, Theorem 1.1]. Let (L,≺) be a partially order set and satisfy that thelengths of chains with the top element u ∈ L in L are bounded from above. Let A be a free

Z[q±12 ]-module spanned by the basis {Eu | u ∈ L}. The bar-involution on A is the Z-linear


map x 7→ x such that fx = fx for f ∈ Z[q±12 ] and x ∈ A, where f(q

12 ) = f(q−

12 ). If

Eu −Eu ∈⊕

u′≺u

Z[q±12 ]Eu′ (4.1)

for u ∈ L, then there exists a unique element Cu ∈ A satisfies that

Cu = Cu, (4.2)

Cu − Eu ∈⊕

u′∈L

q−12Z[q−

12 ]Eu′ . (4.3)

More precisely, (4.3) can be replaced by

Cu − Eu ∈⊕

u′≺u

q−12Z[q−

12 ]Eu′ . (4.4)

Then {Cu | u ∈ L} is a Z[q±12 ]-basis in A.

Let A be the quantum cluster algebra associated with an acyclic quantum seed. By [4,Theorem 1.4, Theorem 1.6], the basis {Cu | u ∈ L} of A is uniquely determined by (4.2)and (4.3). The basis {Cu | u ∈ L} is called the canonical triangular basis in A (see [4,Section 1] for more detail).

For x ∈ Z, the function [x]+ is defined by [x]+ = x if x > 0 and [x]+ = 0 otherwise.In the rest of this section, we set L = Z

2 and A = Aq(1, 4). For (a, b), (a′, b′) ∈ Z

2, recallthat the partial order ≺ on Z

2 used in [4, (6.16)] is defined by:

(a′, b′) ≺ (a, b) ⇐⇒ [−a′]+ < [−a]+, [−b′]+ < [−b]+. (4.5)

Using the same argument as in [4, Section 2], it follows that Aq(1, 4) satisfies the condition(4.1) for the partial order ≺.

The standard monomials in Aq(1, 4) are defined by

E(a,b) = q−12abX

[−a]+3 X

[a]+1 X

[b]+2 X

[−b]+0 (4.6)

for a, b ∈ Z ([4, Section 1]). As in [4], the crystal lattice A+ ⊂ Aq(1, 4) is defined by

A+ =⊕

a,b∈Z

Z[q−12 ]E(a,b).

Recall that in [4] for the the quantum cluster algebra of Kronecker type Aq(2, 2), fora, b ∈ Z, Berenstein and Zelevinsky defined the elements

E′(a,b) := q−

12abX

[−b]+2 X

[b]+0 X

[a]+1 X

[−a]+−1 ∈ Aq(2, 2).

Similarly, we define the elements E′(a,b) and µ1E(a,b) in Aq(1, 4) as follows

E′(a,b) = q−

12abX

[−b]+2 X

[b]+0 X

[a]+1 X

[−a]+−1 and µ1E(a,b) = q−

12abX

[−b]+4 X

[b]+2 X

[a]+3 X

[−a]+1 .

A new partial order � on Z2 is defined by:

(a′, b′) � (a, b) ⇐⇒ [−a′]+ ≤ [−a]+, [−b′]+ ≤ [−b]+.

Remark 4.2. The partial order � is used to prove Lemma 4.3 and Lemma 4.8 and thepartial order ≺ is used in Lusztig’s Lemma.

Lemma 4.3. For a, b ∈ Z, we have that


(1) q12aE(a,b)X0 − E(a,b−1) ∈

⊕

(a′,b′)�(a+1,b−1)

q−12Z[q−

12 ]E(a′,b′);

(2) q12bX3E(a,b) − E(a−1,b) ∈

⊕

(a′,b′)�(a−1,b+4)

q−12Z[q−

12 ]E(a′,b′);

(3) q12bE(a,b)X1 − E(a+1,b) ∈

⊕

(a′,b′)�(a+1,b+4)

q−12Z[q−

12 ]E(a′,b′);

(4) q−12aX4E(a,b) − E(a,b−1) ∈

⊕

(a′,b′)�(a−1,b−1)

q−12Z[q−

12 ]E(a′,b′) if b > 0;

(5) q−12(a−b)X4E(a,b)−E(a−1,b−1) ∈

⊕

(a′,b′)�(a−1,b+3)

q−12Z[q−

12 ]E(a′,b′) if a > 0 and b ≤ 0;

(6) q−12(a−b)X4E(a,b) − E(a−1,b−1) ∈

⊕

(a′,b′)�(a,b)

q−12Z[q−

12 ]E(a′,b′) if a ≤ 0 and b ≤ 0.

Proof. (1) (i) If a ≥ 0, b > 0, then

q12aE(a,b)X0 − E(a,b−1) = q−

12bE(a+1,b−1). (4.7)

(ii) If a ≥ 0, b ≤ 0, then

q12aE(a,b)X0 = q−

12(ab−b)Xa

1X−b+10 = E(a,b−1). (4.8)

(iii) If a < 0, b ≤ 0, then

q12aE(a,b)X0 = q−

12(ab−b)X−a

3 X−b+10 = E(a,b−1). (4.9)

(iv) If a < 0, b > 0, then

q12aE(a,b)X0 −E(a,b−1) = q−

12bE(a+1,b−1) + q−

12(b−4a)E(a+1,b+3). (4.10)

(2) (i) If a ≤ 0, b ≥ 0, then

q12bX3E(a,b) = q−

12(ab−b)X1−a

3 Xb2 = E(a−1,b). (4.11)

(ii) If a ≤ 0, b < 0, then

q12bX3E(a,b) = q−

12(a−1)bX1−a

3 X−b0 = E(a−1,b). (4.12)

(iii) If a > 0, b ≥ 0, then

q12bX3E(a,b) − E(a−1,b) = q−

12(ab+8a−b−4)Xa−1

1 Xb+42 = q−2aE(a−1,b+4). (4.13)

(iv) If a > 0, b = −1, then

q−12X3E(a,−1) − E(a−1,−1) = q−2aE(a−1,3) + q−2a−2E(a,3). (4.14)

(v) If a > 0, b = −2, then

q−1X3E(a,−2) − E(a−1,−2) = q−2aE(a−1,2) + q−2a(q−52 + q−

32 )E(a,2) + q−2a−4E(a+1,2).

(4.15)

(vi) If a > 0, b = −3, then

q−32X3E(a,−3) − E(a−1,−3)

=q−2aE(a−1,1) + q−2a(q−3 + q−2 + q−1)E(a,1) + q−2a(q−5 + q−4 + q−3)E(a+1,1)

+ q−2a−6E(a+2,1). (4.16)


(vii) If a > 0, b ≤ −4, then

q12bX3E(a,b) − E(a−1,b)

=q−2aE(a−1,b+4) + q−12(4a−b−3)(q−3 + q−2 + q−1 + 1)E(a,b+4)

+ q−2a+b+2(q−4 + q−3 + 2q−2 + q−1 + 1)E(a+1,b+4)

+ q−12(4a−3b−3)(q−3 + q−2 + q−1 + 1)E(a+2,b+4) + q−2a+2bE(a+3,b+4). (4.17)

(3) (i) If a ≥ 0, b ≥ 0, then

q12bE(a,b)X1 = q−

12(a+1)bXa+1

1 Xb2 = E(a+1,b). (4.18)

(ii) If a ≥ 0, b < 0, then

q12bE(a,b)X1 = q−

12(a+1)bXa+1

1 X−b0 = E(a+1,b). (4.19)

(iii) If a < 0, b ≥ 0, then

q12bE(a,b)X1 − E(a+1,b) = q2aE(a+1,b+4). (4.20)

(iv) If a = −1, b < 0, then

q12bE(−1,b)X1 − E(0,b) = q−2X4

2X−b0 = q−2E(0,4)X

−b0 .

When a = −1, b = −1, we have that

q−12E(−1,−1)X1 − E(0,−1) = q−2E(0,3) + q−4E(1,3). (4.21)


q−1E(−1,−2)X1 − E(0,−2) = q−2E(0,2) + (q−92 + q−

72 )E(1,2) + q−6E(2,2) (4.22)


q−32E(−1,−3)X1 − E(0,−3)

=q−2E(0,1) + (q−5 + q−4 + q−3)E(1,1) + (q−7 + q−6 + q−5)E(2,1) + q−8E(3,1). (4.23)

When a = −1, b ≤ −4, we have that

q12bE(−1,b)X1 − E(0,b)

=q−2E(0,b+4) + q12b(q−

72 + q−

52 + q−

32 + q−

12 )E(1,b+4)

+ qb(q−4 + q−3 + 2q−2 + q−1 + 1)E(2,b+4) + q32b(q−

72 + q−

52 + q−

32 + q−

12 )E(3,b+4)

+ q2b−2E(4,b+4). (4.24)

(v) If a ≤ −2, b < 0, then

q12bE(a,b)X1 − E(a+1,b) = q−

12(ab+b+4)X−a−1

3 X42X

−b0 = q2a−

12(a+1)bE(a+1,4)X

−b0 .


q−12E(−2,−1)X1 − E(−1,−1) = q−

52X3X

42X0 = q−6E(0,3) + q−8E(0,7) + q−4E(−1,3).


When a ≤ −2, by repeatedly using (4.9) and (4.10), we know that q2aE(a+1,4+b) is a term

of q2a−12(a+1)bE(a+1,4)X

−b0 . We only need to consider the expansion of

q2a−12(a+1)bE(a+1,4)X

−b0 − q2aE(a+1,4+b)

with respect to the standard monomials E(c,d).Note that

E(c,d)X0 = q−12cE(c,d−1) + q−

12(c+d)E(c+1,d−1), if c ≥ 0, d > 0;

E(c,d)X0 = q−12cE(c,d−1), if c ≥ 0, d ≤ 0;

E(c,d)X0 = q−12cE(c,d−1), if c < 0, d ≤ 0;

E(c,d)X0 = q−12cE(c,d−1) + q−

12(c+d)E(c+1,d−1) + q

12(3c−d)E(c+1,d+3), if c < 0, d > 0.

(4.25)

By applying (4.25) repeatedly, we have that

q2a−12(a+1)bE(a+1,4)X

−b0 − q2aE(a+1,4+b) ∈ q

− 12A+

and every term kq−12αE(c,d) (k ∈ N\{0}) in q2a−

12(a+1)bE(a+1,4)X

−b0 −q2aE(a+1,4+b) satisfies

the conditions that α ≥ −4a > 0, c ≥ a+ 1 and d ≥ 4 + b. Namely

q12bE(a,b)X1 − E(a+1,b) = q2a−

12(a+1)bE(a+1,4)X

−b0 ∈

⊕

(a′,b′)�(a+1,b+4)

q−12Z[q−

12 ]E(a′,b′).

(4) (i) If a > 0, b > 0, then

q−12aX4E(a,b) − E(a,b−1) = q−

12bE(a−1,b−1) + q−

12(4a+b)E(a−1,b+3). (4.26)

(ii) If a ≤ 0, b > 0, then

q−12aX4E(a,b) − E(a,b−1) = q−

12bE(a−1,b−1). (4.27)

(5) (i) If a > 0, b = 0, then

q−12aX4E(a,0) − E(a−1,−1) = q−2aE(a−1,3). (4.28)

(ii) If a > 0, b = −1, then

q−12(a+1)X4E(a,−1) − E(a−1,−2) = q−2aE(a−1,2) + q−

12(4a+3)E(a,2). (4.29)

(iii) If a > 0, b = −2, then

q−12(a+2)X4E(a,−2) − E(a−1,−3) = q−2aE(a−1,1) + q−2a(q−2 + q−1)E(a,1) + q−2a−3E(a+1,1).

(4.30)(iv) If a > 0, b ≤ −3, then

q−12(a−b)X4E(a,b) − E(a−1,b−1)

=q−2aE(a−1,b+3) + q−12(4a−b−3)(q−

52 + q−

32 + q−

12 )E(a,b+3)

+ q−12(4a−2b−3)(q−

52 + q−

32 + q−

12 )E(a+1,b+3) + q−

12(4a−3b)E(a+2,b+3). (4.31)


(6) Through a direct calculation, we have that

X4E(0,0) = X4 = E(−1,−1) − q−2E(0,3),

q12X4E(−1,0) = q

12X4X3 = E(−2,−1) − q−4E(−1,3),

q−12X4E(0,−1) = q−

12X4X0 = E(−1,−2) − q−

52E(0,2) − q−4E(1,2).

We will prove the statement by induction on both value a and b. For fixed a ≤ 0, b ≤ 0,assume that

q−12(a−b)X4E(a,b) − E(a−1,b−1) =: ∆ ∈

⊕

(a′,b′)�(a,b)

q−12Z[q−

12 ]E(a′,b′).

Note that

q−12(a−1−b)X4E(a−1,b) = q−

12(1−b)X3q

− 12(a−b)X4E(a,b)

=q−12(1−b)X3E(a−1,b−1) + q−

12(1−b)X3∆ = E(a−2,b−1) + q−

12(1−b)X3∆

and

q−12(a−b+1)X4E(a,b−1) = q−

12(a−b)X4E(a,b)q

− 12(1−a)X0

=q−12(1−a)E(a−1,b−1)X0 + q−

12(1−a)∆X0 = E(a−1,b−2) + q−

12(1−a)∆X0.

Let kq−12αE(c,d) (k ∈ Z \ {0}) be a term in ∆. Note that α + 1 − b + d ≥ α + 1 > 0. By

Lemma 4.3(2), we have that

q−12(1−b)X3q

− 12αE(c,d) = q−

12(α+1−b+d)(q

12dX3E(c,d))

∈q−12(α+1−b+d)(E(c−1,d) +

⊕

(a′,b′)�(c−1,d+4)

q−12Z[q−

12 ]E(a′,b′)) ⊆

⊕

(a′,b′)�(c−1,d)

q−12Z[q−

12 ]E(a′,b′).

It follows that

q−12(a−1−b)X4E(a−1,b) − E(a−2,b−1) ∈

⊕

(a′,b′)�(a−1,b)

q−12Z[q−

12 ]E(a′,b′).

Note that α+ c+ 1− a ≥ α+ 1 > 0. By Lemma 4.3(1), we have that

q−12αE(c,d)q

− 12(1−a)X0 = q−

12(α+c+1−a)(q

12cE(c,d)X0)

∈q−12(α+c+1−a)(E(c,d−1) +

⊕

(a′,b′)�(c+1,d−1)

q−12Z[q−

12 ]E(a′,b′)) ⊆

⊕

(a′,b′)�(c,d−1)

q−12Z[q−

12 ]E(a′,b′).

Thus

q−12(a−b+1)X4E(a,b−1) − E(a−1,b−2) ∈

⊕

(a′,b′)�(a,b−1)

q−12Z[q−

12 ]E(a′,b′).

The proof is completed.�

Lemma 4.4. Let a, b ∈ Z<0 and kq−12αE(c,d) (k ∈ N \ {0}) be a term in q

12bE(a,b)X1 −

E(a+1,b). Then we have that c ≤ a− b+ 1 if d ≥ 0 and c− d ≤ a− b+ 1 if d < 0.


Proof. By (4.21), (4.22), (4.23) and (4.24), it follows that (c, d) ∈ {(0, 3), (1, 3)} if a = −1,b = −1; (c, d) ∈ {(0, 2), (1, 2), (2, 2)} if a = −1, b = −2; (c, d) ∈ {(0, 1), (1, 1), (2, 1), (3, 1)}if a = −1 and b = −3; (c, d) ∈ {(0, b+4), (1, b+4), (2, b+4), (3, b+4), (4, b+4)} if a = −1and b ≤ −4.

When a ≤ −2 and b < 0, q12bE(a,b)X1 −E(a+1,b) = q2a−

12(a+1)bE(a+1,4)X

−b0 . By (4.25), it

follows that c ∈ {a+ 1, . . . , a− b+ 1}, and we obverse that there exists some nonnegativeinteger n ≤ −a− 1 such that

d = 4 + 3n− (−b− n) = 4 + 4n+ b ≤ b− 4a.

If d = 4+4n+ b ≥ 0, then c ≤ (a+1)+n+(−b−n) = a− b+1. If d = 4+4n+ b < 0, thenc ≤ (a+1)+n+(4+3n) = a+5+4n, it follows that c− d ≤ a− b+1. Thus c ≤ a− b+1if d ≥ 0 and c− d ≤ a− b+ 1 if d < 0.

�

By using (4.26) and (4.27), we obtain the following lemma.

Lemma 4.5. For a ∈ Z and b ∈ Z>0, every term q−12αE(c,d) in q−

12aX4E(a,b) − E(a,b−1)

satisfies that c = a− 1 and d ≥ b− 1 ≥ 0.

Lemma 4.6. For a ∈ Z>0 (respectively, a ∈ Z≤0) and b ∈ Z≤0, every term kq−12αE(c,d)

(k ∈ Z \ {0}) in q−12(a−b)X4E(a,b) −E(a−1,b−1) satisfies that c ≥ a− 1 (respectively, c ≥ a),

c ≤ a− b if d ≥ 0 and c− d ≤ a− b if d < 0.

Proof. (1) When a > 0, by using the identities (4.28), (4.29), (4.30) and (4.31), we havethat (c, d) = (a − 1, 3) if b = 0; (c, d) ∈ {(a − 1, 2), (a, 2)} if b = −1; (c, d) ∈ {(a −1, 1), (a, 1), (a+1, 1)} if b = −2; (c, d) ∈ {(a−1, b+3), (a, b+3), (a+1, b+3), (a+2, b+3)}if b ≤ −3. It is easy to see that c ≥ a− 1, c < a− b if d ≥ 0 and c− d < a− b if d < 0.

(2) When a ≤ 0, we will prove the statement by induction. By Lemma 4.3(6), it followsthat c ≥ a. When a = 0, note that

q−12X4E(0,−1) = q−

12X4X0 = E(−1,−2) − q−

52E(0,2) − q−4E(1,2),

q−1X4E(0,−2) = q−1X4X20 = E(−1,−3) − q−3E(0,1) − (q−5 + q−4)E(1,1) − q−6E(2,1),

and

q−12nX4E(0,−n) =E(−1,−n−1) − q−

12(n+4)E(0,3−n) − q−n(q−3 + q−2 + q−1)E(1,3−n)

− q−32n(q−3 + q−2 + q−1)E(2,3−n) − q−2n−2E(3,3−n)

for n ≥ 3. It is easy to see that c ≤ n = a− b if d ≥ 0, c− d ≤ a− b if d < 0.

Note that q12X4E(−1,0) = E(−2,−1) − q−4E(−1,3). When a = −m ≤ −1 and b = 0, we

have that

q−12mX4E(−m,0) = q−

12(m−1)Xm−1

3 (q−12X3X4)

=q−12(m−1)Xm−1

3 E(−2,−1) − q−12(7+m)Xm−1

3 E(−1,3)

=E(−m−1,−1) − q−2(1+m)E(−m,3).

From now on, we assume that a, b ≤ −1 for the rest of the proof. When a = b = −1, wehave that

X4E(−1,−1) = q−1X4X3X0 = E(−2,−2) − q−92E(−1,2) − q−6E(0,2) − q−8E(0,6).


When a = −m, b = −1, we have that

q−1−m

2 X4E(−m,−1) = E(−m−1,−2) − q−4m+5

2 E(−m,2) − q−2m−4E(1−m,2) − q−4m−4E(1−m,6).

For a = −m ∈ Z<0 and b = −n ∈ Z<0, assume that every term kq−12αE(c,d) in

∇ := q−n−m

2 X4E(−m,−n) − E(−m−1,−n−1)

satisfies that c ≤ n−m if d ≤ 0 and c− d ≤ n−m if d < 0.When a = −m and b = −n− 1, we have that

q−12(n+1−m)X4E(−m,−n−1) = q−

12(m+1)(q−

12(n−m)X4E(−m,−n))X0

=q−12(m+1)(E(−m−1,−n−1) +∇)X0 = E(−m−1,−n−2) + q−

12(m+1)∇X0.

Let k′q−12α′

E(c′,d′) (k′ ∈ Z\{0}) be a term in q−

m+12 (kq−

12αE(c,d))X0. When c ≥ 0, d > 0,

it follows that (c′, d′) ∈ {(c, d− 1), (c+ 1, d− 1)} by (4.7), i.e., d′ ≥ 0 and c ≤ c′ ≤ c+ 1 ≤n+1−m. If d ≤ 0 then (c′, d′) = (c, d−1) by (4.8) and (4.9), i.e., c′ = c, d′ < 0 and c′−d′ =c− d+1 ≤ n−m+1. If c < 0, d > 0, then (c′, d′) ∈ {(c, d− 1), (c+1, d+3), (c+1, d− 1)}by (4.10), i.e., d′ ≥ 0 and c ≤ c′ ≤ c+ 1 ≤ n−m+ 1.

The proof is completed. �

By [4, Theorem 3.1, Proposition 4.1], we have the following lemma.

Lemma 4.7. For each (a, b) ∈ Z2, we have that

E′(a,b) − Eϕ(a,b) ∈ q

− 12A+,

where ϕ : Z2 → Z2 defined by ϕ(a, b) = (a,−4[−a]+ − b) is a bijection.

Similarly, let ψ : Z2 → Z2 be a map defined by ψ(a, b) = (−a − [−b]+, b). It is easy to

see that ψ is a bijection.

Lemma 4.8. For each (a, b) ∈ Z2, we have that

µ1E(a,b) − Eψ(a,b) ∈ q−12A+.

Proof. Let a1, a2 ∈ Z≥0, it is clear that

µ1E(a1,a2) = q−12a1a2Xa2

2 Xa13 = E(−a1,a2) and µ1E(−a1,a2) = q−

12a1a2Xa1

1 Xa22 = E(a1,a2).

We need only consider µ1E(a1,−a2) − E(−a1−a2,−a2) and µ1E(−a1,−a2) − E(a1−a2,−a2).(1) When a1 = 1, a2 = 0, we have that µ1E(1,0) = X3 = E(−1,0). When a1 = 0, a2 = 1,

µ1E(0,−1) −E(−1,−1) = X4 − q−12X3X0 = −q−2X3

2 = −q−2E(0,3).

When a1 = 1, a2 = 1, we have that

µ1E(1,−1) − E(−2,−1) = q−12X3X4 − q−1X2

3X0 = −q−112 X3

2X3 = −q−4E(−1,3).

We will prove the statement by induction on both a1 and a2. Assume that

µ1E(a1,−a2) − E(−a1−a2,−a2)

=q−12a1a2Xa1

3 Xa24 − q−

12(a1+a2)a2Xa1+a2

3 Xa20 =: C1 ∈

⊕

(a′,b′)�(−a1−a2,−a2)

q−12Z[q−

12 ]E(a′,b′),

i.e., every term kq−12αE(c,d) (k ∈ Z \ {0}) in C1 satisfies [−c]+ ≤ a1 + a2 and [−d]+ ≤ a2.


(i) By Lemma 4.3(2), we have that

µ1E(a1+1,−a2) = q−12a2X3E(−a1−a2,−a2) + q−

12a2X3C1 = E(−a1−a2−1,−a2) + q−

12a2X3C1

and q−12(α+a2)X3E(c,d) = q−

12(α+a2+d)(q

12dX3E(c,d)). Note that α + a2 + d ≥ α > 0, by

Lemma 4.3(2), we have that q−12(α+a2)X3E(c,d) ∈

⊕

(a′,b′)�(c−1,d)

q−12Z[q−

12 ]E(a′,b′). Thus

µ1E(a1+1,−a2) − E(−a1−a2−1,−a2) ∈⊕

(a′,b′)�(−a1−a2−1,−a2)

q−12Z[q−

12 ]E(a′,b′).

(ii) Note that

µ1E(a1,−a2−1) = qa12 X4(q

− 12a1a2Xa1

3 Xa24 ) = q

a12 X4E(−a1−a2,−a2) + q

a12 X4C1.

By Lemma 4.3(6),

qa12 X4E(−a1−a2,−a2) ∈ E(−a1−a2−1,−a2−1) +

⊕

(a′,b′)�(−a1−a2,−a2)

q−12Z[q−

12 ]E(a′,b′).

When d ≥ 0, we have that q−12(α−a1)X4E(c,d) = q−

12(α−a1−c)(q−

12cX4E(c,d)). When

d < 0, we have that q−12(α−a1)X4E(c,d) = q−

12(α−a1−c+d)(q−

12(c−d)X4E(c,d)). By Lemma

4.6, we know that c ≤ −a1 if d ≥ 0, and c− d ≤ −a1 if d < 0. Thus α− a1 − c ≥ α > 0 ifd ≥ 0, and α− a1 − c+ d ≥ α > 0 if d < 0. By using the induction hypothesis and Lemma4.3(4), (5), (6), we have that

q12a1X4C1 ∈

⊕

(a′,b′)�(−a1−a2−1,−a2−1)

q−12Z[q−

12 ]E(a′,b′).

Hence µ1E(a1,−a2−1) − E(−a1−a2−1,−a2−1) ∈⊕

(a′,b′)�(−a1−a2−1,−a2−1)

q−12Z[q−

12 ]E(a′,b′).

(2) When a1 = 0, a2 > 0, we have that

µ1E(0,−a2) − E(−a2,−a2) ∈⊕

(a′,b′)�(−a2,−a2)

q−12Z[q−

12 ]E(a′,b′)

by (1). When a1 > 0, a2 = 0, we have that µ1E(−a1,0) = Xa11 = E(a1,0).

When a1 = a2 = 1, we have that µ1E(−1,−1) = X0 + q−2X32 = E(0,−1) + q−2E(0,3).

When a1 = 1, a2 = 2, we have that µ1E(−1,−2) − E(−1,−2) = q−4E(−1,2) − q−4E(1,2).

When a1 = 2, a2 = 1, we have that µ1E(−2,−1) − E(1,−1) = q−4E(1,3).When a1 = 2, a2 = 2, we have that

µ1E(−2,−2) − E(0,−2) = (q−6 + q−2)E(0,2) + q−8E(0,6) + (q−92 + q−

72 )E(1,2).

When a1 = 3, a2 = 1, we have that µ1E(−3,−1) − E(2,−1) = q−6E(2,3).When a1 = 3, a2 = 2, we have that

µ1E(−3,−2) − E(1,−2) = (q−8 + q−4)E(1,2) + q−12E(1,6) + (q−132 + q−

112 )E(2,2).

We will proceed the proof by induction on both a1 and a2.


(i) When a1 ≥ a2 ≥ 1, we assume that

µ1E(−a1,−a2) − E(a1−a2,−a2) = q−12a1a2Xa2

4 Xa11 − q−

12(a1−a2)a2Xa2

0 Xa1−a21

=:C2 ∈⊕

(a′,b′)�(0,−a2)

q−12Z[q−

12 ]E(a′,b′)

and every term k1q− 1

2αE(c,d) (k1 ∈ Z \ {0}) in C2 satisfies that c ≥ 0, c ≤ a1 if d ≥ 0 and

c− d ≤ a1 if d < 0. Note that

µ1E(−a1−1,−a2) = q−12a2µ1E(−a1,−a2)X1 = q−

12a2E(a1−a2,−a2)X1 + q−

12a2C2X1

=E(a1+1−a2,−a2) + q−12a2C2X1.

By (4.18) and (4.19), q−12(a2+α)E(c,d)X1 = q−

12(a2+α+d)E(c+1,d).

Note that a2 + α+ d ≥ α > 0 and c+ 1 > c ≥ 0. Thus

q−12a2C2X1 ∈

⊕

(a′,b′)�(0,−a2)

q−12Z[q−

12 ]E(a′,b′) ⊆ q−

12A+.

For a1 ≥ a2 ≥ 1, it follows that

µ1E(−a1−1,−a2) − E(a1+1−a2,−a2) ∈⊕

(a′,b′)�(0,−a2)

q−12Z[q−

12 ]E(a′,b′) ⊆ q−

12A+

and every term k′1q− 1

2α′

E(c′,d′) = k1q− 1

2(a2+α+d)E(c+1,d) in µ1E(−a1−1,−a2) −E(a1+1−a2,−a2)

satisfies that c′ = c+ 1 > 0, c′ ≤ a1 + 1 if d′ = d ≥ 0 and c′ − d′ ≤ a1 + 1 if d′ = d < 0.(ii) When a2 > a1 ≥ 1, we assume that

µ1E(−a1,−a2) − E(a1−a2,−a2)

=q−12a1a2Xa2

4 Xa11 − q−

12(a2−a1)a2Xa2−a1

3 Xa20 =: C3 ∈

⊕

(a′,b′)�(a1−a2,−a2)

q−12Z[q−

12 ]E(a′,b′).


2βE(e,f) (k2 ∈ Z\{0}) in C3 satisfies that e ≤ a1 if f ≥ 0 and e−f ≤ a1

if f < 0. Note that

µ1E(−a1−1,−a2) = q−12a2µ1E(−a1,−a2)X1 = q−

12a2E(a1−a2,−a2)X1 + q−

12a2C3X1.

By Lemma 4.3(3), we obtain that

q−12a2E(a1−a2,−a2)X1 ∈ E(a1+1−a2,−a2) +

⊕

(a′,b′)�(a1+1−a2,4−a2)

q−12Z[q−

12 ]E(a′,b′).

Note that a2 + β + f > 0 since [−f ]+ ≤ a2 and β > 0. Then we have that

q−12(a2+β)E(e,f)X1 ∈ q−

12(a2+β+f)E(e+1,f) +

⊕

(a′,b′)�(e+1,f+4)

q−12Z[q−

12 ]E(a′,b′)

⊆⊕

(a′,b′)�(e+1,f)

q−12Z[q−

12 ]E(a′,b′).

We obtain that q−12a2C3X1 ∈

⊕

(a′,b′)�(a1+1−a2,−a2)

q−12Z[q−

12 ]E(a′,b′).

Let k′2q− 1

2β′

E(e′,f ′) (k2 ∈ Z \ {0}) be a term in µ1E(−a1−1,−a2) − E(a1−a2+1,−a2).


When k′2q− 1

2β′

E(e′,f ′) is a term in q−12a2E(a1−a2,−a2)X1 −E(a1+1−a2,−a2), we obtain that

e′ ≤ a1 + 1 if f ′ ≥ 0 and e′ − f ′ ≤ a1 + 1 if f ′ < 0, by Lemma 4.4.

Now we consider the case that k′2q− 1

2β′

E(e′,f ′) is a term in k2q− 1

2(a2+β)E(e,f)X1, where

k2q− 1

2βE(e,f) is a term in C3. Note that a2 + β + f ≥ β > 0 and

q−12(a2+β)E(e,f)X1 = q−

12(a2+β+f)(q

12fE(e,f)X1).

When e ≥ 0, we have (e′, f ′) = (e+ 1, f) by (4.18) and (4.19). When e < 0 and f ≥ 0, wehave (e′, f ′) = (e+1, f) or (e+1, f +4) by (4.20). When e < 0 and f < 0, we obtain thate′ ≤ e− f + 1 ≤ a1 + 1 if f ′ ≥ 0 and e′ − f ′ ≤ e− f + 1 ≤ a1 + 1 if f ′ < 0 by Lemma 4.4.

Hence

µ1E(−a1−1,−a2) −E(a1−a2+1,−a2) ∈⊕

(a′,b′)�(a1−a2+1,−a2)

q−12Z[q−

12 ]E(a′,b′) ⊆ q−

12A+

for a2 > a1 ≥ 1, and every term k′2q− 1

2β′

E(e′,f ′) in µ1E(−a1−1,−a2) −E(a1+1−a2,−a2) satisfiesthat e′ ≤ a1 + 1 if f ′ ≥ 0, e′ − f ′ ≤ a1 + 1 if f ′ < 0.

(iii) When a1 > a2 ≥ 1, we assume that

µ1E(−a1,−a2) − E(a1−a2,−a2) =: C4 ∈⊕

(a′,b′)�(0,−a2)

q−12Z[q−

12 ]E(a′,b′) ⊆ q−

12A+


2γE(g,h) (k3 ∈ Z \ {0}) in C4 satisfies that g ≥ a1− a2, g ≤ a1 if h ≥ 0

and g − h ≤ a1 if h < 0. We obverse that

µ1E(−a1,−a2−1) = q−12a1X4µ1E(−a1,−a2) = q−

12a1X4E(a1−a2,−a2) + q−

12a1X4C4

and q−12a1X4E(a1−a2,−a2) ∈ E(a1−a2−1,−a2−1) +

⊕

(a′,b′)�(0,3−a2)

q−12Z[q−

12 ]E(a′,b′) by Lemma

4.3(5).Similar to the proof of Case (2)(ii), according to Lemma 4.5 and Lemma 4.6, we can

prove that every term k′3q− 1

2γ′E(g′,h′) in µ1E(−a1,−a2−1) − E(a1−a2−1,−a2−1) satisfies the

conditions that g′ ≥ a1 − a2 − 1 ≥ 0, g′ ≤ a1 if h′ ≥ 0, g′ − h′ ≤ a1 if h′ < 0 and

µ1E(−a1,−a2−1) − E(a1−a2−1,−a2−1) ∈⊕

(a′,b′)�(0,−a2−1)

q−12Z[q−

12 ]E(a′,b′) ⊆ q−

12A+

for a1 > a2 ≥ 1.(iv) When a2 ≥ a1 ≥ 1, we assume that

µ1E(−a1,−a2) − E(a1−a2,−a2) = q−12a1a2Xa2

4 Xa11 − q−

12(a2−a1)a2Xa2−a1

3 Xa20

=:C5 ∈⊕

(a′,b′)�(a1−a2,−a2)

q−12Z[q−

12 ]E(a′,b′) ⊆ q−

12A+


2δE(i,j) (k4 ∈ Z\{0}) in C5 satisfies that i ≤ a1 if j ≥ 0 and i− j ≤ a1

if j < 0. Then

µ1E(−a1,−a2−1) = q−12a1X4µ1E(−a1,−a2) = q−

12a1X4E(a1−a2,−a2) + q−

12a1X4C5


and q−12a1X4E(a1−a2,−a2) ∈ E(a1−a2−1,−a2−1)+

⊕

(a′,b′)�(a1−a2,−a2)

q−12Z[q−

12 ]E(a′,b′) by Lemma

4.3(6).

Let k′4q− 1

2δ′E(i′,j′) (k

′4 ∈ Z\{0}) be a term in µ1E(−a1,−a2−1)−E(a1−a2−1,−a2−1). Similar

to the proof of Case (2)(ii), according to Lemma 4.3(5), (6), Lemma 4.5 and Lemma 4.6,it follows that

q−12a1X4C5 ∈

⊕

(a′,b′)�(a1−a2−1,−a2−1)

q−12Z[q−

12 ]E(a′,b′)


2δ′E(i′,j′) in q

− 12a1X4C5 satisfies that i′ ≤ a1 if j′ ≥ 0 and i′ − j′ ≤ a1

if j′ < 0.In a summary, we obtain that

µ1E(−a1,−a2−1) − E(a1−a2−1,−a2−1) ∈⊕

(a′,b′)�(a1−a2−1,−a2−1)

q−12Z[q−

12 ]E(a′,b′) ⊆ q−

12A+


2δ′E(i′,j′) in µ1E(−a1,−a2−1) −E(a1−a2−1,−a2−1) satisfies that i

′ ≤ a1 ifj′ ≥ 0 and i′ − j′ ≤ a1 if j′ < 0.

This completes the proof of Lemma 4.8. �

For n ∈ Z, the notation α(n) is defined by

〈n〉α(n) =

{

(2− n, 2(3 − n)), if n ≥ 2;

(n, 2(n − 1)), if n ≤ 1.

Recall that C(a,b) is the element of the canonical triangular basis for any a, b ∈ Z.

Proposition 4.9. For n ∈ Z and a1, a2 ∈ Z≥0, we have

Ca1α(n)+a2α(n+1) = q−12a1a2Xa1

n Xa2n+1.

Proof. Through a direct calculation, we obtain that σ−2(µ1E(a,b)) = E′(a,b). Recall that ϕ

and ψ are bijective. By Lemma 4.8, µ1Eψ(a,b)−E(a,b) ∈ q−12A+. ThusE

′ψ(a,b)−σ−2(E(a,b)) ∈

q−12A+. By Lemma 4.7, we obtain that

Eϕψ(a,b) − σ−2(E(a,b)) ∈ q−12A+. (4.32)

By Lemma 4.8, E′(a,b) − σ−2(Eψ(a,b)) ∈ q−

12A+. It follows that Eϕ(a,b) − σ−2(Eψ(a,b)) ∈

q−12A+ and E(a,b) − σ−2(Eψϕ(a,b)) ∈ q−

12A+. Hence

σ2(E(a,b))− Eψϕ(a,b) ∈ q−12A+. (4.33)

Using (4.32) and (4.33), it follows that

σ−2(C(a,b)) = Cϕψ(a,b) = C(−a−[−b]+,−4[a+[−b]+]+−b) (4.34)

and

σ2(C(a,b)) = Cψϕ(a,b) = C(−a−[4[−a]++b]+,−4[−a]+−b). (4.35)


According to the definition of the triangular basis, it follows that

C(a1,a2) = E(a1,a2) = q−12a1a2Xa1

1 Xa22 ;

C(a1,−a2) = E(a1,−a2) = q−12a1a2Xa2

0 Xa11 ;

C(−a1,a2) = E(−a1,a2) = q−12a1a2Xa2

2 Xa13

(4.36)

for a1, a2 ∈ Z≥0

From (4.34), (4.35) and (4.36), it is deduced that Ca1α(n)+a2α(n+1) = q−12a1a2Xa1

n Xa2n+1

for a1, a2 ∈ Z≥0 and n ∈ Z.�

Now we need only consider C(−n,−2n) for positive integers n.

Lemma 4.10. If n is a positive integer, then

Sn(Xδ) =q−nX

〈n〉n+2X

20 − q−(n+2)X

〈n+1〉n+1 X1 − q−(n+1)(q−

12 + q

12 )

⌊n2⌋+1

∑

k=1

X〈n+1〉n+3−2k

−∑

k≥1

kq−2k(q−12 + q

12 )2Sn−2k(Xδ). (4.37)

Proof. We will prove the lemma by induction on n. When n = 1, by [6, Lemma 3.6], wehave that

Xδ = q−1X3X20 − q−3X2

2X1 − q−2(q−12 + q

12 )X2

2 .

When n is even, by the induction hypothesis,

Sn+1(Xδ) = XδSn(Xδ)− Sn−1(Xδ)

=q−nXδX2n+2X

20 − q−(n+2)XδXn+1X1 − q−(n+1)(q−

12 + q

12 )

n2+1

∑

k=1

XδXn+3−2k

−∑

k≥1

kq−2k(q−12 + q

12 )2XδSn−2k(Xδ)− q−(n−1)Xn+1X

20 + q−(n+1)X2

nX1

+ q−n(q−12 + q

12 )

n2

∑

k=1

X2n+2−2k +

∑

k≥1

kq−2k(q−12 + q

12 )2Sn−1−2k(Xδ).

A direct calculation shows that

q−nXδX2n+2X

20 = q1−nXn+1X

20 + q−n−1Xn+3X

20 + q−n(q−

12 + q

12 )X2

0 ,

q−(n+2)XδXn+1X1 = q−(n+1)X2nX1 + q−(n+3)X2

n+2X1,

XδXn+3−2k = qXn+2−2k + q−1Xn+4−2k.

Note that

q−n(q−12 + q

12 )

n2+1

∑

k=1

X2n+2−2k = q−n(q−

12 + q

12 )

n2

∑

k=1

X2n+2−2k + q−n(q−

12 + q

12 )X2

0 .


Thus

Sn+1(Xδ) =q−(n+1)Xn+3X

20 − q−(n+3)X2

n+2X1 − q−(n+2)(q−12 + q

12 )

n2+1

∑

k=1

X2n+4−2k

−∑

k≥1

kq−2k(q−12 + q

12 )2Sn+1−2k(Xδ).

When n is odd, by the induction hypothesis,

Sn+1(Xδ) =XδSn(Xδ)− Sn−1(Xδ)

=q−nXδXn+2X20 − q−(n+2)XδX

2n+1X1 − q−(n+1)(q−

12 + q

12 )

n+12

∑

k=1

XδX2n+3−2k

−∑

k≥1

kq−2k(q−12 + q

12 )2XδSn−2k(Xδ)− q−(n−1)X2

n+1X20 + q−(n+1)XnX1

+ q−n(q−12 + q

12 )

n+12

∑

k=1

Xn+2−2k +∑

k≥1

kq−2k(q−12 + q

12 )2Sn−1−2k(Xδ).

We have that

q−nXδXn+2X20 = q−(n−1)X2

n+1X20 + q−(n+1)X2

n+3X20 ,

q−(n+2)XδX2n+1X1 = q−(n+1)XnX1 + q−(n+3)Xn+2X1 + q−(n+2)(q−

12 + q

12 )X1,

q−(n+1)(q−12 + q

12 )

n+12

∑

k=1

XδX2n+3−2k

=q−n(q−12 + q

12 )

n+12

∑

k=1

Xn+2−2k + q−(n+2)(q−12 + q

12 )

n+12

∑

k=1

Xn+4−2k +n+ 1

2q−(n+1)(q−

12 + q

12 )2

and

∑

k≥1

kq−2k(q−12 + q

12 )2

(

XδSn−2k(Xδ)− Sn−1−2k(Xδ))

=

n−12

∑

k=1

kq−2k(q−12 + q

12 )2Sn+1−2k(Xδ).

Hence we obtain that

Sn+1(Xδ) =q−(n+1)X2

n+3X20 − q−(n+3)Xn+2X1 − q−(n+2)(q−

12 + q

12 )

n+32

∑

k=1

Xn+4−2k

−∑

k≥1

kq−2k(q−12 + q

12 )2Sn+1−2k(Xδ).

The proof is completed. �

Lemma 4.11. For n ∈ Z≥0, we have that C(−n,−2n) = Sn(Xδ).


Proof. We will prove the statement by induction on n. It is trivial for n = 0. For n = 1,by Theorem 3.3, we have that

E(−1,−2) = Xδ + (q−52 + q−

32 )E(0,2) + q−4E(1,2).

Therefore C(−1,−2) = Xδ. Assume that C(−r,−2r) = Sr(Xδ) for 0 ≤ r ≤ n. When n is even,we have that

Sn(Xδ) =q−nX2

n+2X20 − q−(n+2)Xn+1X1 − q−(n+1)(q−

12 + q

12 )

n2+1

∑

k=1

Xn+3−2k

−∑

k≥1

kq−2k(q−12 + q

12 )2Sn−2k(Xδ).

By Proposition 4.9, we know that X2n+2 = C2α(n+2) = C(−n,2−2n) and Xn+1 = Cα(n+1) =

C(1−n,4−2n). By using Lemma 4.3(1), (3) and the condition (4.4) in Lusztig’s Lemma, weobtain that

q−nC(−n,2−2n)X20 ∈ q−nE(−n,2−2n)X

20 + q−

12A+

and

q−(n+2)C(1−n,4−2n)X1 ∈ q−(n+2)E(1−n,4−2n)X1 + q−12A+.

For n ≥ 2, q−nE(−n,2−2n)X20 = q−

n2 (q−

n2E(−n,2−2n)X0)X0 = q−

n2E(−n,1−2n)X0 = E(−n,−2n)

and q−(n+2)E(1−n,4−2n)X1 ∈ q−3(E(2−n,4−2n) + q−12A+) ⊆ q−

12A+ by (4.9) and Lemma

4.3(3). Hence we have that

q−nX2n+2X

20 = q−nC(−n,2−2n)X

20 ∈ E(−n,−2n) + q−

12A+

and

q−(n+2)Xn+1X1 = q−(n+2)C(−n,4−2n)X1 ∈ q−12A+.

For 1 ≤ k ≤ n2 + 1, it is clearly that

q−(n+1)(q−12 + q

12 )Xn+3−2k

=q−(n+1)(q−12 + q

12 )Cα(n+3−2k) ∈ q−(n+1)(q−

12 + q

12 )(E(2k−n−1,4k−2n) + q−

12A+) ⊆ q−

12A+.

By using the induction hypothesis, we have that q−2k(q−12 + q

12 )2Sn−2k(Xδ) ∈ q−

12A+ for

1 ≤ k ≤ n2 . When n is even, we have that

Sn(Xδ) ∈ E(−n,−2n) + q−12A+.

When n is odd, we have that

Sn(Xδ) =q−nXn+2X

20 − q−(n+2)X2

n+1X1 − q−(n+1)(q−12 + q

12 )

n+12

∑

k=1

X2n+3−2k

−∑

k≥1

kq−2k(q−12 + q

12 )2Sn−2k(Xδ),

Xn+2 = Cα(n+2) = C(−n,2−2n), X2n+1 = C2α(n+2) = C(1−n,4−2n) andX

2n+3−2k = C2α(n+3−2k)

= C(2k−n−1,4k−2n) for 1 ≤ k ≤ n+12 . The rest of the proof is quite similar to that given

above for the case of n being even and so is omitted.


Since Sn(Xδ) is bar-invariant, we are led to the conclusion that C(−n,−2n) = Sn(Xδ) forn ∈ Z≥0. �

Proposition 4.9 and Lemma 4.11 imply the following theorem.

Theorem 4.12. The basis S = {q−12abXa

mXbm+1 | m ∈ Z, (a, b) ∈ Z

2≥0} ∪ {Sn(Xδ)} is the

triangular basis in Aq(1, 4).

Acknowledgments

The first draft of the paper was written during X. Chen’s visit at Nankai Universityfrom May 19 to June 18, 2017. He thanks Nankai University for the hospitality and forcreating ideal working environment. We appreciate Prof. Fan Qin’s comments on therelation between the bases obtained in our paper and some other existed bases.

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Department of Applied Mathematics, Northwestern Polytechnical University, Xi’an, Shaanxi

710072, P. R. China

E-mail address: [email protected] (L.Bai)

Department of Mathematics, University of Wisconsin-Whitewater, 800 W. Main Street,

Whitewater, WI.53190. USA

E-mail address: [email protected] (X.Chen)

School of Mathematical Sciences and LPMC, Nankai University, Tianjin, P. R. China

E-mail address: [email protected] (M.Ding)

Department of Mathematical Sciences, Tsinghua University, Beijing 100084, P. R. China

E-mail address: [email protected](F.Xu)

http://arxiv.org/abs/1606.05604

arxiv:1706.04376v1 [math.qa] 14 jun 2017 · 2 liqianbai,xueqingchen,mingdingandfanxu extended the...

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